Question

Integrate:

$\int \frac{1}{3+2\sin x}dx$

Answer 1

Let,

$I=\int \frac{1}{2\sin x+3}dx$

$I=\int \frac{1}{\frac{4\tan \left(\frac{x}{2}\right)}{{\tan }^2\left(\frac{x}{2}\right)+1}+3}dx$

Substitute,

$u=\tan \frac{x}{2}$

$dx=\frac{2}{{\sec }^2\left(\frac{x}{2}\right)}du=\frac{2}{u^2+1}du$

So,

$I=2\int \frac{1}{3u^2+4u+3}du$

$I=2\int \frac{1}{{(\sqrt{3}u+\frac{2}{\sqrt{3}})}^2+\frac{5}{3}}du$

Substitute,

$v=\frac{3u+2}{\sqrt{5}}$

$\Rightarrow du=\frac{\sqrt{5}}{3}dv$

Therefore,

$I=2\int \frac{\sqrt{5}}{5v^2+5}dv$

$I=\frac{2}{\sqrt{5}}\int \frac{1}{v^2+1}dv$

$I=\frac{2}{\sqrt{5}}{\tan }^{-1}\left(v\right)$

$I=\frac{2}{\sqrt{5}}{\tan }^{-1}\left(\frac{3u+2}{\sqrt{5}}\right)$

$I=\frac{2}{\sqrt{5}}{\tan }^{-1}\left(\frac{3\tan \frac{x}{2}+2}{\sqrt{5}}\right)+C$

Hence, this is the required result.